one liner creative solution for Just Fizz! by viktor.chyrkin - python coding challenges

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one liner solution in Creative category for Just Fizz! by viktor.chyrkin

checkio = lambda v: 'Fizz' if v % 3 == 0 else str(v)

Nov. 28, 2022

Comments:

mu_py on Dec. 20, 2022, 9:32 p.m.
Could you please explain, why you consider this solution to be "creative"? To me it looks quite straightforward. Oh, and please forgive my nosiness, but... Why do you check for `== 0`, instead of writing `checkio = lambda v: str(v) if v % 3 else 'Fizz'` ?

viktor.chyrkin on Dec. 22, 2022, 7:37 p.m.
I just felt like putting it in creative that's all.

viktor.chyrkin on Dec. 22, 2022, 7:38 p.m.
And for <pre class='brush: python'>==0 </pre> I am just used to doing it that way.

Strings and Integers

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